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Practice Exam 4 Solution

# Practice Exam 4 Solution - Math 323 Exam 4 Practice...

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Math 323 Exam 4 - Practice Problems 1. Evaluate ii R ( y + x ) dA , where R is the region bounded by x = 0, y = 0, and 2 x + y = 4. -1 -2 -1 1 2 3 4 5 6 7 1 2 3 4 5 6 -2 -3 x y R i 2 0 i 4 2 x 0 y + xdy dx = i 2 0 1 2 y 2 + xy v v v v v 4 2 x 0 dx = i 2 0 1 2 (4 - 2 x ) 2 + x (4 - 2 x ) - 0 dx = i 2 0 1 2 (16 - 16 x + 4 x 2 ) + 4 x - 2 x 2 dx = i 2 0 8 - 4 xdx = 8 x - 2 x 2 v v v v v 2 0 = 16 - 8 = 8 2. For each of the double integrals given below, ±rst graph the region of integration. Then reverse the order of integration for the iterated integral and then evaluate the integral exactly. (a) i 1 0 i 1 y 3 xe x 3 dxdy x=1 2 1 2 1 y x R x=y = i 1 0 i x 0 3 xe x 3 dy dx = i 1 0 3 xye x 3 v v v v v x 0 dx = i 1 0 3 x 2 e x 3 dx = e x 3 v v v v v 1 0 = e 1 - e 0 = e - 1 (b) i 1 0 i 1 x 3 4 + y 3 dy dx 2 y=1 1 2 1 x y = x y R = i 1 0 i y 2 0 3 4 + y 3 = i 1 0 3 x 4 + y 3 v v v v v y 2 0 dy = i 1 0 3 y 2 4 + y 3 dy = ln | 4 + y 3 | v v v v v 1 0 = ln 5 - ln 4 = ln 5 4

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(c) i π 0 i π y p 1 - cos x x P dxdy x 3 2 1 2 3 1 x x=y R y = i π 0 i x 0 1 - cos x x dy dx = i π 0 y - y cos x x v v v v v x 0 dx = i π 0 x - cos xdx = 1 2 x - sin x v v v v v π 0 = π 2 2 - 0 = π 2 2 3. Express the volume of the solid bounded by the curves z = x + 2 ,z = 0 ,x = y 2 - 2 and x = y as a double integral in rectangular coordinates. Also, sketch the region in the plane for the integration. DO NOT EVALUATE THE INTEGRAL. -1 -2 -2 -1 2 x = y 1 2 1 x 2 x = y - 2 y R Notice that the points of intersection for the region in the plane occur when y = y 2 - 2, or when y 2 - y - 2 = 0. That is, when ( y - 2)( y + 1) = 0, so when y = 2 or y = - 1. Therefore, the volume is given by: i 2 1 i y y 2 2 x + 2 4. Evaluate ii R xdA where R is the region in the polar plane bounded by r = 1 - sin θ . K 1.0 K 0.5 0 0.5 1.0 K 2.0 K 1.5 K 1.0 K 0.5 Converting everything to polar coodinates, x = r cos θ and r = 1 - sin θ is the cardioid graphed above. Then we have i 2 π 0 i 1 sin θ 0 r cos θr dr dθ = i 2 π 0 i 1 sin θ 0 r 2 cos θ dr dθ = i 2 π 0 1 3 r 2 cos θ v v v v v 1 sin θ 0 = 1 3 i 2 π 0 (1 - sin θ ) 3 cos θ dθ = 1 3 i 2 π 0 (1 - 3 sin θ + 3 sin 2 θ - sin 3 θ ) cos θ dθ = 1 3 sin θ - 3 2 sin 2 θ + sin 3 θ - 1 4 sin 4 θ v v v v v 2 π 0 = 0 [Note that this is not a surprise as f ( x,y ) = x is odd and the region R is symmetric with respect to the y -axis]
5. Calculate the mass of a lamina that occupies the plane region R bounded by the curve ( x - 1) 2 + y 2 = 1 with density function ρ ( x,y ) = 1 r x 2 + y 2 .

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Practice Exam 4 Solution - Math 323 Exam 4 Practice...

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